Chapter 1 Logic of Compound Statements

UCCM1333 INTRODUCTORY DISCRETE MATHEMATICS
Chapter 1 Logic of Compound Statements
Statements and Logical form
Definition 1.1
A statement or proposition is a declarative sentence that is either true or
false, but not both.

Definition 1.2
The truth value of a proposition is true (T), if it is a true proposition and
false (F), if it is a false proposition.

Example 1.1
p: The year 1973 was a leap year.
is a proposition readily decidable as false.

Note the use of the label ‘p: …..’, so that the overall statement is read ‘p
is the statement: “The year 1973 was a leap year” ’. So we use p, q, r, s
and t to represent statements and these letters are called statement
variables, that is, variables that can be replaced by statements.

Example 1.2
Determine whether the following sentences are statements or not. If it is a
statement, determine its truth value.

(a) Selangor is a state in Malaysia.

(b) The sun rises in the West.

(c) She is a computer science major.

(d) 128 = 26.

(e) x = 26.

(f) Is (210 –1) an even integer?

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(g) Take the book.

(h) x2 + x + 1 = 0, x is a real number.

(i) x2 + x + 1 = 0, x is a complex number.

(j) Maths is fun

Definition 1.3
A table that gives the truth values of a compound statement in terms of its
component parts is called a truth table.

Connectives
Most mathematical statements are combinations of simpler statement
formed through some choice of the words not, and, or, if___then___,
and if and only if. These are called logical connectives (or simply
connectives) and are denoted by the following symbols:
∼ or ¬ Not
∧ And
∨ Or
⇒ If ____, then____
⇔ If and only if

Compound statements
Definition 1.4
(1) A statement represented by a single statement variable (without
any connective) is called a simple (or primitive) statement.
(2) A statement represented by some combination of statement
variables and connectives is called a compound statement.

Example 1.3
(1) A dog or a car is an animal.
(2) A dog is not an animal.
(3) 5 < 3.
(4) If the earth is flat, then 3 + 4 = 7.

2

Truth table
A truth table displays the relationships between the truth values of
statements. Truth tables are especially valuable in the determination of
the truth values of statements constructed from simpler statements

Definition 1.5
If p is a statement variable, the negation of p is “not p” or “it is not the
case that p” and is denoted ∼p. It has opposite truth value from p.

Truth Table for ∼p
p ∼p
T F
F T

Example 1.4
Give the negation of the following statements:
(a) p: The integer 10 is even.
(b) q: 2 + 3 > 1
(c) r: 3 + 7 = 10

Definition 1.6
If p and q are statement variables, the conjunction of p and q is “p and
q”, denoted p ∧ q. The compound statement p ∧ q is true when both p and
q are true; otherwise, it is false.
Truth Table for p ∧ q
p q p∧q
T T T
T F F
F T F
F F F

Example 1.5
Form the conjunction of p and q for each of the following:
(a) p: It is snowing. q: I am cold

(b) p: 2 < 3. q: −5 > –8

3

Example 1.6
Determine the truth value of the following statements:
(a) 3 < 5 and 5 + 6 ≠ 11

(b) 5 is positive and Kuala Lumpur is in Malaysia.

(c) The integer 2 is even but it is a prime number.

Definition 1.6
If p and q are statement variables, the disjunction of p and q is “p or q”,
denoted p ∨ q. The compound statement p ∨ q is true if at least one of p
or q is true; it is false when both p and q are false.
Truth Table for p ∨ q
p q p∨q
T T T
T F T
F T T
F F F

Note
The notation for inequalities involves “and” and “or” statements:
Let a, b and c be particular real numbers.
a≤b means a < b or a = b
a<b<c means a < b and b < c.

Note
∼ is an unary operation, while ∨ and ∧ are binary operations.

Example 1.7
Form the disjunction of p and q for the following:
p: 2 is a positive integer. q: 2 is a rational number.

Example 2.8
Determine the truth value of the given statements:
(a) 3 or −5 is negative.
(b) 2 or π is an integer.
(c) 5 ≤ 5

4

Definition 1.7
A statement form is an expression made up of statement variables and
connectives.
The truth table for a given statement form displays the truth values that
correspond to the different combinations of truth values for the variables.

Note
If a statement form s contains n statement variables, there will need to be
2n rows in the truth table for s.

Example 1.9
s: p ∨ (q ∧ (p ∨ r)) involves 3 statement variables, p, q and r. So there are
altogether 23 or 8 possible combinations of truth values for p, q and r.

Truth Table for s (s involves 2 statement variables)
p q s
T T
T F
F T
F F

Truth Table for s (s involves 3 statement variables)
p q r s
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

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Example 1.10
Construct a truth table for the statement form (p ∧ q) ∨ ∼r.
Steps:
(1) Set up columns labeled p, q, r, ∼r, (p ∧ q) and (p ∧ q) ∨ ∼r.
(2) Fill in the p, q and r columns with all the logically possible
combinations of T’s and F’s.
(3) Use the truth tables for ∼ and ∧ to fill in the ∼r and (p ∧ q)
columns with the appropriate truth values.
(4) Finally, fill in the (p ∧ q) ∨ ∼r column by considering truth values
for (p ∧ q) and ∼r.

Truth table for (p ∧ q) ∨ ∼r
p q r ∼r (p ∧ q) (p ∧ q) ∨ ∼r
T T T F T T
T T F T T T
T F T F F F
T F F T F T
F T T F F F
F T F T F T
F F T F F F
F F F T F T

Example 1.11
Construct a truth table for the following statement forms:
(i) (∼p ∨ q) ∧ (∼r)
(ii) (∼p ∧ ∼r) ∨ (r ∧ q)
(iii) q ∧ ∼(∼p ∨ r)

(i) Truth table for (∼p ∨ q) ∧ (∼r)
p q r
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

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(ii) Truth table for (∼p ∧ ∼r) ∨ (r ∧ q)
p q r
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

(iii) Truth table for q ∧ ∼(∼p ∨ r)
p q r
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

Tautology, Contradiction and Contingency
Definition 1.8
A tautology is a statement form where its truth values in all rows in the
truth table are always true.
A contradiction is a statement form where its truth values in all rows in
the truth table are always false.
A contingency is a statement form that is neither tautology nor
contradiction.

Note
Normally, t is used to denote a tautology and c is used to denote a
contradiction.

Example 1.12
Let p, q and r be statement variables. Show that the statement form
(a) ∼p ∨ p is a tautology
(b) ∼p ∧ p is a contradiction
(c) (p ∧ q) ∨ ∼r is a contingency.
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(a) & (b)

(c)

Conditional Statements
Definition 1.9
If p and q are statements, the statement “if p then q” or “p implies q”,
denoted p ⇒ q, is called a conditional statement, or implication.
The statement p is called the hypothesis and the statement q is called the
conclusion (or consequent).
It is false, when p is true and q is false; otherwise it is true.
Truth Table p ⇒ q
p q p⇒q
T T T
T F F
F T T
F F T

Note
(1) A variety of terminology is used to express p ⇒ q as given below:
If p then q q if p
p implies q q when p
p only if q q follows from p
p is sufficient for q q is necessary for p

(2) If the hypothesis p is false, we consider p ⇒ q is true regardless of
its conclusion.
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p ⇒ q is true because of p is false is often called true by default.
A true conditional statement does not mean a conditional statement
with true conclusion.

Example 1.13
Determine the hypothesis and conclusion for each of the following
conditional statements. Then determine the truth value.
(1) The moon is square only if the sun rises in the East.

(2) 1 and 3 are prime if 1 multiply 3 is prime.

(3) (sin π)(cos π) = 0 when sin π = 0 or cos π = 0.

(4) If 1 + 1 = 3, then cats can fly.

Negation, Contrapositive, Converse and Inverse
Definition 1.10
Let p and q be statement variables:
(1) The negation of p ⇒ q is p ∧ ∼q.
(2) The contrapositive of p ⇒ q is ∼q ⇒ ∼p.
(3) The converse of p ⇒ q is q ⇒ p.
(4) The inverse of p ⇒ q is ∼p ⇒ ∼q.

Note
Note that the truth tables of p ⇒ q and its contrapositive are the same,
and the truth tables of the converse and the inverse are the same.

Example 1.14
Write the negation, contrapositive, converse and inverse of the following
conditional statements:
(a) If 3 is positive then 3 is nonnegative.
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Negation:

Contrapositive:

Converse:

Inverse:

(b) If you study hard, then you will not fail UTSM3363.
Negation:

Contrapositive:

Converse:

Inverse:

Only if
To say “p only if q” means that if q does not take place, then p cannot
take place.

Definition 1.11
Let p and q be statement variables, p only if q means “if not q then not p”,
or equivalently “if p then q”.

Biconditional
Definition 1.12
Let p and q be statement variables. The statement form
(p ⇒ q) ∧ (q ⇒ p)
Is called the biconditional of p and q.
This is read “p if and only if q” and is denoted p ⇔ q.
It is true if both p and q have the same truth values and it is false if p and
q have the opposite truth values.
Truth Table for p ⇔ q
p q p⇔q
T T T
T F F
F T F
F F T
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Note
p is a necessary and sufficient condition for q means p ⇔ q .

Example 1.15
Determine the truth values of the following statements:
(a) 2 is prime if and only if it is multiple of 2.

(b) 2 is negative if and only if 4 is negative.

(c) 0<1⇔2<1

Note
(1) The order of operations can be overridden through the use of
parentheses – ( ).
(2) The symbol ∧ and ∨ are considered coequal in order of operation.
(3) The order of operations is that ∼ is performed first, then ∧ and ∨,
and finally ⇒ and ⇔.

Exclusive Or
Definition 1.13
The connective corresponding to the exclusive or is denoted by ⊕.
p ⊕ q is true when exactly one of p and q is true.

Note
p ⊕ q and p ⇔ q have opposite truth values.

Logical Equivalence
Definition 1.14
Two statement forms P and Q are called logically equivalence if, and
only if P ⇔ Q is a tautology.
The logical equivalence of statement forms P and Q is written as P ≡ Q.

Note
One way to determine whether 2 statement forms P and Q are logically
equivalent is to construct truth table P ⇔ Q.

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Example 1.16
Let p and q be 2 statement variables. Determine whether the following
statement forms are logically equivalent or not.
(a) ∼p ∨ ∼q and ∼(p ∨ q)

(b) p ∨ (p ∧ q) and p

Negations of conjunction and disjunction: De Morgan’s Law
De Morgan’s Law
∼(p ∧ q) ≡ ∼p ∨ ∼q
∼(p ∨ q) ≡ ∼p ∧ ∼q

Note
∼(p ∧ q) is called the negation of conjunction of p and q.
∼(p ∨ q) is called the negation of disjunction of p and q.

Example 1.17
Write the negation of the given statements:
(1) John is smart but lazy.

(2) 2 ≤2< π

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Logical equivalences of conditional statements
Theorem 1.1
Let p and q be statements variables.
(1) p ⇒ q ≡ ∼p ∨ q
(2) p ⇒ q ≡ ∼q ⇒ ∼p
(3) p ⇔ q ≡ (p ⇒ q) ∧ (q ⇒ p)

Example 1.18
Prove the logical equivalences
(1) p ⇒ q ≡ ∼p ∨ q (2) p ⇒ q ≡ ∼q ⇒ ∼p

Laws of Logical Equivalences
Theorem 1.2
Given any statement variables p, q and r, a tautology t and a
contradiction c, the following logical equivalences hold.
(1) Commutative laws: p∧q≡q∧p
p∨q≡q∨p

(2) Associative laws: (p ∧ q ) ∧ r ≡ p ∧ (q ∧ r )
(p ∨ q ) ∨ r ≡ p ∨ (q ∨ r )

(3) Distributive laws: p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(4) Identity laws: p∧t≡p
p∨c≡p

(5) Negation laws: p ∨ ∼p ≡ t
p ∧ ∼p ≡ c
(6) Double negative laws: ∼(∼p) ≡ p
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(7) Idempotent laws: p∧p≡p
p∨p≡p

(8) Universal bound laws: p∨t≡t
p∧c≡c

(9) De Morgan’s laws: ∼(p ∧ q) ≡ ∼p ∨ ∼q
∼(p ∨ q) ≡ ∼p ∧ ∼q

(10) Absorption laws: p ∨ (p ∧ q) ≡ p
p ∧ (p ∨ q) ≡ p

(11) Negations of t and c: ∼t ≡ c
∼c ≡ t

Example 1.19
Show that
(a) (p ∨ q) ∧ ∼(∼p ∧ q) ≡ p
(b) ∼[∼((p ∨ q) ∧ r) ∨ ∼ q] ≡ q ∧ r

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Example 1.20
Simplify the following statement, stating the law used in each step of the
simplification:
(p ∨ q) ∧ p ∧ (r ∨ q) ∧ (p ∨ ∼p ∨ r) ∧ (r ∨ ∼q)

Rules of Inferences
Definition 1.15
An argument is a sequence of statements written
p1
p2
M or p1, p2, …, pn / ∴ q (∴is read as therefore)
pn
_______
∴ q
which means p1 ∧ p2 ∧ …∧ pn ⇒ q.
The statements p1, p2, …, pn are called called premises (or hypothesis or
assumptions) and the statement q is called conclusion.

Definition 1.16
An argument is valid provided that if p1 and p2 and …and pn are all true,
then q must be also true; otherwise the argument is invalid (or fallacy).
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Testing an argument form for validity
(1) Identity the premises and conclusion
(2) Construct a truth table showing the truth values of all the premises
and the conclusion.
(3) If the truth table contains a row in which all premises are true and
the conclusion is false, then the argument form is invalid.
Otherwise, in every case where all the premises are true, the
conclusion is also true, then the argument form is valid.

Example 1.21
Determine the validity of the following argument forms by using truth
tables.
(i) p ⇒ q ∨ ∼r
q⇒p∧r
∴ p⇒r

p q r ∼r q ∨ ∼r p∧r p ⇒ q ∨ ∼r q⇒p∧r p⇒r
T T T F T T T T T
T T F T T F T F F
T F T F F T F T F
T F F T T F T T F
F T T F T F T F F
F T F T T F T F F
F F T F F F T T T
F F F T T F T T T

(ii) p ∨ (q ∨ r)
∼r
∴ p∨q

p q r p ∨ (q ∨ r) ∼r p∨q
T T T T F T
T T F T T T
T F T T F T
T F F T T T
F T T T F T
F T F T T T
F F T T F F
F F F F T F
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Rules of Inference (some valid argument forms)
Rules of Inference Name
1 p⇒q
p Modus Ponens
__________ (Law of Detachment)
∴q

2 p⇒q
∼q Modus Tollens
__________
∴∼p

3 p or q
________ ________ Disjunctive Addition
∴p∨q ∴p∨q

4 p∧q or p∧q
________ ________ Conjunctive Simplification
∴p ∴q

5 p∨q or p∨q
∼p ∼q Disjunctive Syllogism
________ _________
∴q ∴p

6 p
q Conjunction
________
∴p∧q

7 p⇒q
q⇒r Hypothetical Syllogism
________
∴p⇒r

8 p∨q
∼p ∨ r Resolution
________
∴q∨r

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Example 1.23
Use rules of inferences to fill in the blanks in the following arguments so
as to produce valid inferences.

(a) If a figure is a square, then it is a rectangle.
S is a square.

(b) If I have finished my assignment, I will send you an email.

∴ I didn’t finish my assignment

(c) I will enroll in Math I or Physic I in Year 2.
I didn’t enroll in Physic I in Year 2

Example 1.24
We are given the following:
If the Charges get a good linebacker, then the Charges can beat Broncos.
If the Charges can beat the Broncos, then the Chargers can beat the Jets.
If the Chargers can beat the Broncos, then the Chargers can beat the
Dolphins. The Chargers get a good linebacker.
Show by using the rules of inference that the conclusion, the Chargers
can beat the Jets and the Chargers can beat the Dolphins, follows from
the hypotheses.

Let
p = the Charges get a good linebacker
q = the Charges can beat Broncos
r = the Charges can beat the Jets
s = the Charges can beat the Dolphins

Show conclusion: r ∧ s

Step Reason
(1) p⇒q hypothesis
(2) q⇒r hypothesis
(3) q⇒s hypothesis
(4) p hypothesis

(5) p⇒r from (1) & (2), we use hypothetical syllogism
18

(6) r from (4) & (5), we use modus ponens
(7) p⇒s from (1) & (3), we use hypothetical syllogism

(8) s from (4) & (7), we use modus ponens
(9) r∧s from (6) & (8), we use conjunction
So we conclude that the conclusion does follow from the hypotheses

Example 1.25
Show that the hypotheses
“If John takes the computer course, then John stays in the hostel”
“John does not stay in the hostel”
“If John does not take the computer course, then John takes the language
course or stay at home”
“If John takes language course then John buys a motorcycle”
“If John buys a car, then John does not buy motorcycle”
“John has a car”
lead to the conclusion “John stays at home”
Let
p = John takes the computer course
q = John stays in the hostel
r = John takes the language course
s = John stays at home
t = John buys a motorcycle
u = John buys a car

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Converse Error
Definition 1.17
Modus Ponens states that ((p ⇒ q) ∧ p) ⇒ q.
The argument ((p ⇒ q) ∧ q) ⇒ p is invalid and is called the converse
error because the conclusion of the argument would follow from the
premises if the premises p ⇒ q were replaced by its converse. Such a
replacement is not allowed, however, because a conditional statement is
not logically equivalent to its converse.

Inverse Error
Definition 1.18
Modus Tollens states that ((p ⇒ q) ∧ ∼q) ⇒ ∼p.
The argument ((p ⇒ q) ∧ ∼ p) ⇒ ∼q is invalid and is called the inverse
error because the conclusion of the argument would follow from the
premises if the premise p ⇒ q were replaced by its inverse. Such a
replacement is not allowed, however, because a conditional statement is
not logically equivalent to its inverse.

Example 1.26
Are the following arguments valid?
(1) If Ali is tall, then he sits in the back row.
Ali sits in the back row.
Therefore Ali is tall

(2) If interest rates are going up, stock market prices will go down.
Interest rates are not going up.
Therefore stock market prices will not go down.

Example 1.27
Determine the validity of the following arguments:
(a) p⇒q
∼p ⇒ r
r⇒s
∴ ∼q ⇒ s

(b) (∼p ∨ ∼q) ⇒ (r ∧ s)
r⇒t
∼t
∴ p
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Chapter 1 Logic of Compound Statements

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